Answer
(a) $\rho = 2.34\times 10^{17}~kg/m^3$
(b) $r = 12.7~km$
Work Step by Step
(a) $^{238}U$ has 238 nucleons.
We can assume that the nucleus is a sphere.
We can find the density of the nucleus:
$\rho = \frac{m}{V}$
$\rho = \frac{(238)(1.67\times 10^{-27}~kg)}{\frac{4}{3}\pi~r^3}$
$\rho = \frac{(3)(238)(1.67\times 10^{-27}~kg)}{(4\pi)~(7.4\times 10^{-15}~m)^3}$
$\rho = 2.34\times 10^{17}~kg/m^3$
(b) We can find the radius of the neutron star:
$\rho = \frac{m}{V}$
$V = \frac{m}{\rho}$
$\frac{4}{3}\pi~r^3 = \frac{m}{\rho}$
$r^3 = \frac{3m}{4\pi~\rho}$
$r = \sqrt[3] {\frac{3m}{4\pi~\rho}}$
$r = \sqrt[3] {\frac{(3)(2.0\times 10^{30}~kg)}{(4\pi)(2.34\times 10^{17}~kg/m^3)}}$
$r = 1.27\times 10^4~m$
$r = 12.7~km$
